Molecular Geometry and Electron Domain Theory

Molecular Geometry and Electron Domain Theory

Module by: John S. Hutchinson





Foundation

We begin by assuming a
Lewis structure model for chemical bonding based on
valence shell electron pair sharing and the octet rule. We thus
assume the nuclear structure of the atom, and we further assume the
existence of a valence shell of electrons in each atom which
dominates the chemical behavior of that atom. A covalent chemical
bond is formed when the two bonded atoms share a pair of valence
shell electrons between them. In general, atoms of Groups IV
through VII bond so as to complete an octet of valence shell
electrons. A number of atoms, including C, N, O, P, and S, can form
double or triple bonds as needed to complete an octet. We know that
double bonds are generally stronger and have shorter lengths than
single bonds, and triple bonds are stronger and shorter than double
bonds.

Goals

We should expect that the properties of
molecules, and correspondingly the substances which they comprise,
should depend on the details of the structure and bonding in these
molecules. The relationship between bonding, structure, and
properties is comparatively simple in
diatomic molecules, which contain two atoms only,
e.g. HClHCl
or
O2O2.
A
polyatomic molecule contains more than two atoms. An
example of the complexities which arise with polyatomic molecules
is molecular geometry: how are the atoms in the molecule arranged
with respect to one another? In a diatomic molecule, only a single
molecular geometry is possible since the two atoms must lie on a
line. However, with a triatomic molecule (three atoms), there are
two possible geometries: the atoms may lie on a line, producing a
linear molecule, or not, producing a bent molecule. In molecules
with more than three atoms, there are many more possible
geometries. What geometries are actually observed? What determines
which geometry will be observed in a particular molecule? We seek a
model which allows us to understand the observed geometries of
molecules and thus to predict these geometries.

Once we have developed an understanding of the
relationship between molecular structure and chemical bonding, we
can attempt an understanding of the relationship of he structure
and bonding in a polyatomic molecule to the physical and chemical
properties we observe for those molecules.

Observation 1: Geometries of molecules

The geometry of a molecule includes a
description of the arrangements of the atoms in the molecule. At a
simple level, the molecular structure tells us which atoms are
bonded to which. At a more detailed level, the geometry includes
the lengths of all of these bonds, that is, the distances between
the atoms which are bonded together, and the angles between pairs
of bonds. For example, we find that in water,
H2OH2O,
the two hydrogens are bonded to the oxygen and each O-H bond length
is 95.72pm (where
1pm=10-12m1pm10-12m).
Furthermore,
H2OH2O
is a bent molecule, with the H-O-H angle equal to 104.5°.
(The measurement of these geometric properties is difficult,
involving the measurement of the frequencies at which the molecule
rotates in the gas phase. In molecules in crystalline form, the
geometry of the molecule is revealed by irradiating the crystal
with x-rays and analyzing the patterns formed as the x-rays
diffract off of the crystal.)

Not all triatomic molecules are bent, however.
As a common example,
CO2CO2
is a linear molecule. Larger polyatomics can have a variety of
shapes, as illustrated in Figure 1.
Ammonia,
NH3NH3,
is a pyramid-shaped molecule, with the hydrogens in an equilateral
triangle, the nitrogen above the plane of this triangle, and a
H-N-H angle equal to 107°. The geometry of
CH4CH4
is that of a tetrahedron, with all H-C-H angles equal to
109.5°. (See also Figure 2(a).)
Ethane,
C2H6C2H6,
has a geometry related to that of methane. The two carbons are
bonded together, and each is bonded to three hydrogens. Each H-C-H
angle is 109.5° and each H-C-C angle is
109.5°. By contrast, in ethene,
C2H4C2H4,
each H-C-H bond angle is 116.6° and each H-C-C bond angle is
121.7°. All six atoms of ethene lie in the same plane. Thus,
ethene and ethane have very different geometries, despite the
similarities in their molecular formulae.
Figure 1Molecular Structures

Molecular Structures (fig1.png)


We begin our analysis of these geometries by
noting that, in the molecules listed above which do
not contain double or triple bonds
(H2OH2O,
NH3NH3,
CH4CH4and
C2H6C2H6),
the bond angles are very similar, each equal to or very close to
the tetrahedral angle 109.5°. To account for the observed
angle, we begin with our valence shell electron pair sharing model,
and we note that, in the Lewis structures of these molecules, the
central atom in each bond angle of these molecules contains four pairs
of valence shell electrons. For methane and ethane, these four
electron pairs are all shared with adjacent bonded atoms, whereas
in

ammonia or water, one or two (respectively) of
the electron pairs are not shared with any other atom. These
unshared electron pairs are called
lone pairs . Notice that, in the two molecules with no
lone pairs, all bond angles are
exactly equal to the tetrahedral angle, whereas
the bond angles are only close in the molecules with lone
pairs

One way to understand this result is based on
the mutual repulsion of the negative charges on the valence shell
electrons. Although the two electrons in each bonding pair must
remain relatively close together in order to form the bond,
different pairs of electrons should arrange themselves in such a
way that the distances between the pairs are as large as possible.
Focusing for the moment on methane, the four pairs of electrons
must be equivalent to one another, since the four C-H bonds are
equivalent, so we can assume that the electron pairs are all the
same distance from the central carbon atom. How can we position
four electron pairs at a fixed distance from the central atom but
as far apart from one another as possible? A little reflection
reveals that this question is equivalent to asking how to place
four points on the surface of a sphere spread out from each other
as far apart as possible. A bit of experimentation reveals that
these four points must sit at the corners of a tetrahedron, an
equilateral triangular pyramid, as may be seen in Figure 2(b). If the carbon atom is at the
center of this tetrahedron and the four electron pairs at placed at
the corners, then the hydrogen atoms also form a tetrahedron about
the carbon. This is, as illustrated in Figure 2(a), the correct geometry of a methane
molecule. The angle formed by any two corners of a tetrahedron and
the central atom is 109.5°, exactly in agreement with the
observed angle in methane. This model also works well in predicting
the bond angles in ethane.
Figure 2Tetrahedral Structure of Methane
(a) The dotted lines illustrate that the hydrogens form a tetrahedron about the carbon atom.

Figure 2(a) (fig2a.png)

(b) The same tetrahedron is formed by placing four points on a sphere as far apart from one another as possible.

Figure 2(b) (fig2b.png)


We conclude that molecular geometry is
determined by minimizing the mutual repulsion of the valence shell
electron pairs. As such, this model of molecular geometry is often
referred to as the
valence shell electron pair repulsion (VSEPR) theory .
For reasons that will become clear, extension of this model implies
that a better name is the
Electron Domain (ED) Theory .

This model also accounts, at least
approximately, for the bond angles of
H2OH2O
and
NH3NH3.
These molecules are clearly not tetrahedral, like
CH4CH4,
since neither contains the requisite five atoms to form the
tetrahedron. However, each molecule does contain a central atom
surrounded by four pairs of valence shell electrons. We expect from
our Electron Domain model that those four pairs should be arrayed
in a tetrahedron, without regard to whether they are bonding or
lone-pair electrons. Then attaching the hydrogens (two for oxygen,
three for nitrogen) produces a prediction of bond angles of
109.5°, very close indeed to the observed angles of
104.5° in
H2OH2O
and 107° in
NH3NH3.

Note, however, that we do not describe the
geometries of
H2OH2O
and
NH3NH3
as "tetrahedral," since the
atoms of the molecules do not form
tetrahedrons, even if the valence shell electron pairs do. (It is
worth noting that these angles are not exactly equal to
109.5°, as in methane. These deviations will be discussed
later.)

We have developed the Electron Domain model to
this point only for geometries of molecules with four pairs of
valence shell electrons. However, there are a great variety of
molecules in which atoms from Period 3 and beyond can have more
than an octet of valence electrons. We consider two such molecules
illustrated in Figure 3.
Figure 3More Molecular Structures

More Molecular Structures (fig3.png)


First,
PCl5PCl5
is a stable gaseous compound in which the five chlorine atoms are
each bonded to the phosphorous atom. Experiments reveal that the
geometry of
PCl5PCl5
is that of a
trigonal bipyramid : three of the chlorine atoms form
an equilateral triangle with the P atom in the center, and the
other two chlorine atoms are on top of and below the P atom. Thus
there must be 10 valence shell electrons around the phosphorous
atom. Hence, phosphorous exhibits what is called an
expanded valence in
PCl5PCl5.
Applying our Electron Domain model, we expect the five valence
shell electron pairs to spread out optimally to minimize their
repulsions. The required geometry can again be found by trying to
place five points on the surface of a sphere with maximum distances
amongst these points. A little experimentation reveals that this
can be achieved by placing the five points to form a trigonal
bipyramid. Hence, Electron Domain theory accounts for the geometry
of
PCl5PCl5.

Second,
SF6SF6 is
a fairly unreactive gaseous compound in which all six fluorine
atoms are bonded to the central sulfur atom. Again, it is clear
that the octet rule is violated by the sulfur atom, which must
therefore have an expanded valence. The observed geometry of
SF6SF6,
as shown in Figure 3, is highly
symmetric: all bond lengths are identical and all bond angles are
90°. The F atoms form an
octahedron about the central S atom: four of the F
atoms form a square with the S atom at the center, and the other
two F atoms are above and below the S atom. To apply our Electron
Domain model to understand this geometry, we must place six points,
representing the six electron pairs about the central S atom, on
the surface of a sphere with maximum distances between the points.
The requisite geometry is found, in fact, to be that of an
octahedron, in agreement with the observed geometry.

As an example of a molecule with an atom with
less than an octet of valence shell electrons, we consider boron
trichloride,
BCl3BCl3.
The geometry of
BCl3BCl3 is
also given in Figure 3: it is
trigonal planar , with all four atoms lying in the same
plane, and all Cl-B-Cl bond angles equal to 120°. The three Cl
atoms form an equilateral triangle. The Boron atom has only three
pairs of valence shell electrons in
BCl3BCl3.
In applying Electron Domain theory to understand this geometry, we
must place three points on the surface of a sphere with maximum
distance between points. We find that the three points form an
equilateral triangle in a plane with the center of the sphere, so
Electron Domain is again in accord with the observed
geometry.

We conclude from these predictions and
observations that the Electron Domain model is a reasonably
accurate way to understand molecular geometries, even in molecules
which violate the octet rule.

Observation 2: Molecules with Double or Triple Bonds

In each of the molecules considered up to this
point, the electron pairs are either in single bonds or in lone
pairs. In current form, the Electron Domain model does
not account for the observed geometry of
C2H4C2H4,
in which each H-C-H bond angle is 116.6° and each H-C-C bond
angle is 121.7° and all six atoms lie in the same plane.
Each carbon atom in this molecule is surrounded by four pairs of
electrons, all of which are involved in bonding,
i.e. there are no lone pairs. However, the
arrangement of these electron pairs, and thus the bonded atoms,
about each carbon is not even approximately tetrahedral. Rather,
the H-C-H and H-C-C bond angles are much closer to 120°, the
angle which would be expected if
three electron pairs were separated in the
optimal arrangement, as just discussed for
BCl3BCl3.

This observed geometry can be understood by
re-examining the Lewis structure. Recall that, although there are
four electron pairs about each carbon atom, two of these pairs form
a double bond between the carbon atoms. It is tempting to assume
that these four electron pairs are forced apart to form a
tetrahedron as in previous molecules. However, if this were this
case, the two pairs involved in the double bond would be separated
by an angle of 109.5° which would make it impossible for
both pairs to be localized between the carbon atoms. To preserve
the double bond, we must assume that the two electron pairs in the
double bond remain in the same vicinity. Given this assumption,
separating the three
independent groups of electron pairs about a
carbon atom produces an expectation that all three pairs should lie
in the same plane as the carbon atom, separated by 120°
angles. This agrees very closely with the observed bond angles. We
conclude that the our model can be extended to understanding the
geometries of molecules with double (or triple) bonds by treating
the multiple bond as two electron pairs confined to a single
domain. It is for this reason that we refer to the
model as Electron Domain theory.

Applied in this form, Electron Domain theory
can help us understand the linear geometry of
CO2CO2.
Again, there are four electron pairs in the valence shell of the
carbon atom, but these are grouped into only two domains of two
electron pairs each, corresponding to the two C=O double bonds.
Minimizing the repulsion between these two domains forces the
oxygen atoms to directly opposite sides of the carbon, producing a
linear molecule. Similar reasoning using Electron Domain theory as
applied to triple bonds correctly predicts that acetylene,
HCCHHCCH,
is a linear molecule. If the electron pairs in the triple bond are
treated as a single domain, then each carbon atom has only two
domains each. Forcing these domains to opposite sides from one
another accurately predicts 180° H-C-C bond angles.

Observation 3: Distortions from Expected Geometries

It is interesting to note that some molecular
geometries
(CH4CH4,
CO2CO2,
HCCHHCCH)
are exactly predicted by the Electron Domain model, whereas in
other molecules, the model predictions are only approximately
correct. For examples, the observed angles in ammonia and water
each differ slightly from the tetrahedral angle. Here again, there
are four pairs of valence shell electrons about the central atoms.
As such, it is reasonable to conclude that the bond angles are
determined by the mutual repulsion of these electron pairs, and are
thus expected to be 109.5°, which is close but not
exact.

One clue as to a possible reason for the
discrepancy is that the bond angles in ammonia and water are both
less than 109.5°. Another is that both
ammonia and water molecules have lone pair electrons, whereas there
are no lone pairs in a methane molecule, for which the Electron
Domain prediction is exact. Moreover, the bond angle in water, with
two lone pairs, is less than the bond angles in ammonia, with a
single lone pair. We can straightforwardly conclude from these
observations that the lone pairs of electrons must produce a
greater repulsive effect than do the bonded pairs. Thus, in
ammonia, the three bonded pairs of electrons are forced together
slightly compared to those in methane, due to the greater repulsive
effect of the lone pair. Likewise, in water, the two bonded pairs
of electrons are even further forced together by the two lone pairs
of electrons.

This model accounts for the comparative bond
angles observed experimentally in these molecules. The valence
shell electron pairs repel one another, establishing the geometry
in which the energy of their interaction is minimized. Lone pair
electrons apparently generate a greater repulsion, thus slightly
reducing the angles between the bonded pairs of electrons. Although
this model accounts for the observed geometries, why should lone
pair electrons generate a greater repulsive effect? We must guess
at a qualitative answer to this question, since we have no
description at this point for where the valence shell electron
pairs actually are or what it means to share an electron pair. We
can assume, however, that a pair of electrons shared by two atoms
must be located somewhere between the two nuclei, otherwise our
concept of "sharing" is quite meaningless. Therefore, the powerful
tendency of the two electrons in the pair to repel one another must
be significantly offset by the localization of these electrons
between the two nuclei which share them. By contrast, a lone pair
of electrons need not be so localized, since there is no second
nucleus to draw them into the same vicinity. Thus more free to move
about the central atom, these lone pair electrons must have a more
significant repulsive effect on the other pairs of
electrons.

These ideas can be extended by more closely
examining the geometry of ethene,
C2H4C2H4
. Recall that each H-C-H bond angle is 116.6° and each H-C-C
bond angle is 121.7°, whereas the Electron Domain theory
prediction is for bond angles exactly equal to 120°. We can
understand why the H-C-H bond angle is slightly less than
120° by assuming that the two pairs of electrons in the C=C
double bond produce a greater repulsive effect than do either of
the single pairs of electrons in the C-H single bonds. The result
of this greater repulsion is a slight "pinching" of the H-C-H bond
angle to less than 120°.

The concept that lone pair electrons produce a
greater repulsive effect than do bonded pairs can be used to
understand other interesting molecular geometries. Sulfur
tetrafluoride,
SF4SF4,
is a particularly interesting example, shown in Figure 4.
Figure 4Molecular Structure of SF4

Molecular Structure of SF4 (fig4.png)


Note that two of the fluorines form close to a
straight line with the central sulfur atom, but the other two are
approximately perpendicular to the first two and at an angle of
101.5° to each other. Viewed sideways, this structure
looks something like a seesaw.

To account for this structure, we first
prepare a Lewis structure. We find that each fluorine atom is
singly bonded to the sulfur atom, and that there is a lone pair of
electrons on the sulfur. Thus, with five electron pairs around the
central atom, we expect the electrons to arrange themselves in a
trigonal bipyramid, similar to the arrangement in
PCl5PCl5
in Figure 3. In this case, however,
the fluorine atoms and the lone pair could be arranged in two
different ways with two different resultant molecular structures.
The lone pair can either go on the axis of the trigonal
bipyramid (i.e. “above” the sulfur) or on the
equator of the bipyramid (i.e. “beside” the sulfur).

The actual molecular structure in Figure 4 shows clearly that the lone pair
goes on the equatorial position. This can be understood if we
assume that the lone pair produces a greater repulsive effect than
do the bonded pairs. With this assumption, we can deduce that the
lone pair should be placed in the trigonal bipyramidal arrangement
as far as possible from the bonded pairs. The equatorial position
does a better job of this, since only two bonding pairs of
electrons are at approximately 90° angles from the
lone pair in this position. By contrast, a lone pair in the axial
position is approximately 90° away from three bonding
pairs. Therefore, our Electron Domain model assumptions are
consistent with the observed geometry of
SF4SF4.
Note that these assumptions also correctly predict the observed
distortions away from the 180° and
120° angles which would be predicted by a trigonal
bipyramidal arrangement of the five electron pairs.

Review and Discussion Questions
Exercise 1

Using a styrofoam or rubber ball, prove to
yourself that a tetrahedral arrangement provides the maximum
separation of four points on the surface of the ball. Repeat this
argument to find the expected arrangements for two, three, five,
and six points on the surface of the ball.
Exercise 2

Explain why arranging points on the surface of
a sphere can be considered equivalent to arranging electron pairs
about a central atom.
Exercise 3

The valence shell electron pairs about the
central atom in each of the molecules
H2OH2O,
NH3NH3,
and
CH4CH4 are
arranged approximately in a tetrahedron. However, only
CH4CH4 is
considered a tetrahedral molecule. Explain why these statements are
not inconsistent.
Exercise 4

Explain how a comparison of the geometries of
H2OH2O
and
CH4CH4 leads
to a conclusion that lone pair electrons produce a greater
repulsive effect than do bonded pairs of electrons. Give a physical
reason why this might be expected.
Exercise 5

Explain why the octet of electrons about each
carbon atom in ethene,
C2H4C2H4,
are not arranged even approximately in a tetrahedron.
Exercise 6

Assess the accuracy of the following reasoning
and conclusions:

A trigonal bipyramid forms when there are five
electron domains. If one ED is a lone pair, then the lone pair
takes an equatorial position and the molecule has a seesaw
geometry. If two EDs are lone pairs, we have to decide among the
following options: both axial, both equatorial, or one axial and
one equatorial. By placing both lone pairs in the axial positions,
the lone pairs are as far apart as possible, so the trigonal planar
structure is favored.


Exercise 7

Assess the accuracy of the following reasoning
and conclusions:

The Cl-X-Cl bond angles in the two molecules are
identical, because the bond angle is determined by the repulsion of
the two Cl atoms, which is identical in the two molecules.



Figure 5
Figure 5 (fig5.png)

From: http://cnx.org/content/m12594/latest/